150kg cart is traveling right 3m/s. a 20kg block of coal comes from the cart at 38degrees below the horizontal. The new velocity of the cart with the block is 3.95m/s. What is the initial speed of the coal.

the cart speeded up so I assume the coal came AT the cart from behind

cos 38 = .788
so
s cos 38 = .788 s = u = horizontal speed of coal

20 u + 150 (3) = 170 (3.95)

20 u + 450 = 671.5

u = 11.08
s = 11.08/cos 38 = 14.05 m/s

Thanks Damon

To find the initial speed of the coal, we can apply the principle of conservation of momentum. The momentum before the coal is thrown is equal to the momentum after the coal is thrown.

The momentum before the coal is thrown can be calculated as the product of the mass of the cart (150 kg) and its initial velocity (3 m/s):

Momentum_before = mass_of_cart * initial_velocity_of_cart
= 150 kg * 3 m/s
= 450 kg·m/s

The momentum after the coal is thrown can be calculated as the sum of the momentum of the cart and the momentum of the coal:

Momentum_after = momentum_of_cart + momentum_of_coal

Since the coal is thrown horizontally, the vertical component of its momentum is zero. Therefore, the momentum of the coal can be calculated as the product of its mass (20 kg) and its initial horizontal velocity (v):

Momentum_of_coal = mass_of_coal * initial_velocity_of_coal

Now, we need to calculate the horizontal component of the momentum of the cart after the coal is thrown. We can use basic trigonometry to find this component:

horizontal_component_of_cart_after = mass_of_cart * final_velocity_of_cart * cos(theta)
= 150 kg * 3.95 m/s * cos(38 degrees)

Now, equating the momentum before and after the coal is thrown, we have:

Momentum_before = Momentum_after
450 kg·m/s = (mass_of_cart * final_velocity_of_cart * cos(theta)) + (mass_of_coal * initial_velocity_of_coal)

Substituting the given values into the equation, we have:

450 kg·m/s = (150 kg * 3.95 m/s * cos(38 degrees)) + (20 kg * initial_velocity_of_coal)

Now, we can solve for the initial velocity of the coal:

initial_velocity_of_coal = (450 kg·m/s - (150 kg * 3.95 m/s * cos(38 degrees))) / 20 kg

Solving this equation will give us the initial velocity of the coal.